I came across this exercise that says, Show that for any constant $c$, $$u(t) = ca^t \frac{\Gamma(t-t_1)\Gamma(t-t_2) \cdots\Gamma(t-t_n) }{\Gamma(t-s_1)\Gamma(t-s_2) \cdots\Gamma(t-s_m)}$$ is a solution of the recurrence relation$$u(t+1) = a \frac{(t-t_1)(t-t_2) \cdots(t-t_n) }{(t-s_1)(t-s_2) \cdots(t-s_m)}u(t)$$ where $a,t_1,\cdots,t_n,s_1,\cdots ,s_m$ are real constants and $n,m \in \mathbb{N}$. Use this to solve the difference equation
$$\Delta u = \frac{4t + 6}{t^2 + 5t + 6}u$$ where $t \in \mathbb{N}$
The first part of showing that u(t) is a solution I directly substituted $t$ by $t+1$ in the formular of $u(t)$ and got the reccurrence relation. Is this correct?
Now to solve the difference equation here is what I tried:
$\Delta u = u(t+1) - u(t) = \frac{4t + 6}{t^2 + 5t + 6}u$ which implies
$u(t+1) = \frac{t^2 + 9t + 12}{(t+3)(t+2)}u$ and I am stuck since I can not put the numerator in the factorized form inorder to use the relation given above. How do I proceed from this step?
EDIT: I found(on the online errata of the book) the author had made a typing error in the question the equation was supposed to be $$\Delta u = -\frac{4t + 6}{t^2 + 5t + 6}u$$ which becomes easy to solve